I can t figure out two math problems for half a day

1 Let the distance between A and B be x kilometers.

then the prescribed time is (x 4) 9 hours.

x/12+20/60=(x-4)/9

Solution: x=28

The prescribed time is 8 3 hours (160 minutes), distance: 28 km.

2 (1) The two options are the same when purchasing x sticks.

200 20 (x 20) 40 (20 200 40x) solution x 100

2) Scheme a: 20 200 (50 20) 40 5200b scheme: (20 200 50 40).

Plan A saves money.

Dizzy, before October 17th, it was already over.

The landlord is here to have fun.

1) (x-4)/9=x/12+1/3

x = 28 km Stipulated time = 8 3 hours.

2) There is a problem with the second question, a shop owner is now going to the garment factory to buy 20 suits and AThere is a contradiction between buying a suit and getting a tie, and the question is how many ties are bought, isn't it a store owner, not buying 20 suits?

Don't understand!

x-4) 9=(x+20*12 60) 12x=28 km The allotted time is (x-4) 9=2 hours and 40 minutes. Solution: Set up a tie x strip.

1) 200 * 20 + 40 * (x-20) = 20 * x = 100 records.

2) Scheme a; 200 * 20 + (50-20) * 40 = 5200 yuan.

Option b; 200* yuan.

A: Plan A saves money.

A can not transport a box, so A has a combination of 5 out of 2 = 10 kinds, of which the combination containing B is 4 kinds, and the combination without b is 6 kinds, 1) when A selects a container containing B, B has a combination of 4 out of 2, that is, 6 kinds, at this time there are a total of 4 * 6 = 24 kinds.

2) When A selects a container without B, B has a combination of 3 out of 2, that is, 3 kinds, at this time, 6 * 3 = 18 kinds.

So there are 24 + 18 = 42 species, 4, 1There are no restrictions, and the allocation scheme is c(2,6)*c(2,4)*c(2,2) 90

2.If truck A transports A and B is not restricted, the distribution scheme is c(1,5)*c(2,4)*c(2,2) 30

3.If truck B transports B and A is not restricted, the distribution scheme is c(1,5)*c(2,4)*c(2,2) 30

4.If truck A transports A and truck B transports B, the distribution scheme is c(1,4)*c(1,3)*c(2,2) 12

Combined, 1-2-3+4 is the result. 2. There are 2*3*3 kinds when A is not lucky.

There are 4*2*3 kinds when A is transported b.

As soon as the two are added together, it comes out.

It's equal to 42,1, I can't figure out a math problem, please help you.

ABCDEF has a total of 6 containers, ready to be transported by three trucks, two per truck. If A can't ship box A, B can't ship box B, and there are no other restrictions, how many different distribution schemes are there?

Help a friend ask how to answer, I heard that the answer is 42, please help me.

If you extend the length of a rectangular brother by 5 centimeters, the area will increase by 60 square centimeters. If the width is shortened by 5 centimeters, the area is reduced by 100 square centimeters. Please help me calculate the original area of this rectangle.

Solution: The length is extended, and the width remains the same.

I shorten the width and keep the length unchanged.

So the width of the rectangle is.

60 5 = 12 (cm).

The length is: 100 5 = 20 (cm).

Therefore, the area of the rectangle is:

12 20 = 240 (square centimeters).

A: The area of this rectangular envy is 240 square centimeters.

I don't know if the answer you're looking for is this:

The original loose width: 60 5 = 12 (cm).

Wonju lifted length: 100 5 = 20 (cm).

Original area: 20 12 = 240 (square centimeters).

1. If the distance of the expressway between A and B is x kilometers, there are: (x 4-30) 7 x+30

Solution: x = 320 km.

2x-3）=1-2x

Solution: x=7 6

8-k=2(x+1) yields: x=3-k2

Because the solutions are the same, so: 7 6 = 3-k 2

So k = 11 3

The issue has been resolved.

Collection. ** to QQ space.

There are two math problems that won't.

Label: Math problems.

1 drove from A to B originally needed 7 hours, after the opening of the expressway, the distance was nearly 30 kilometers, and the average speed increased by 30 meters per hour, only 4 hours to arrive, find the distance between A and B on the highway.

2.The solution of equations 2(2x-3)=1-2x and 8-k=2(x+1) is the same when k is taken.

Write clearly during the problem-solving process.

Solve with equations. I'll give you the best. Languid.

Satisfied with the answer. 1: Solution: Let the distance of the expressway between A and B be x kilometers.

x+30）/7=x/4-30

4(x+30)=7x-30*28

7x-4x=840+120

x = 320 km.

2：2（2x-3）=1-2x

4x-6=1-2x

4x+2x=1+6

x=7 6 substitution to 8-k=2(x+1).

8-k=2*13*6

k=11/3

The bottom of 48 is 12, so the sum of the top three sides is 12

The second side on the left is 3, so if you cut 3 on the second side on the right side and the right side is 9, then the sum of the remaining sides on the left and the second side on the right is 9, and the answer is 2*(3+9+12)=48

Applying panning results in a rectangle that is 12 long and 9 wide, but has 2 extra 3cm

So the circumference is (12 + 9) 2 + 2 3 = 48cm (give points, it's not easy to get a big night).

（12+9）x2+3x2=48

Piece the shape together from scratch into a square and two 3cm line segments.

Anyway, the first time I saw it, I didn't know how to do it, and the way I read the books should be about the grade, it seems that I am really weak!!

Imagine moving the first and second lines of the upper line upwards and on the same line as the third line, the length of this line is 12

If the third vertical line counted from the left is moved to the far left, which coincides with the first vertical line, the length of the first vertical line is 9

Circumference: 9 + 12 + 9 + 12 + 3 (the second vertical line) + 3 (the part of the first vertical line that coincides) = 48

Count all horizontal 12 12

Count all vertical 9, 9, 3, 3

Add up 24 24 48

It's not a 12-long rectangle with a width of 9, it should have a 12 on both sides, a 9 on one side and a 7 on the other

So c = 12 + 12 + 9 + 7 = 40

48, just forget it by translating the line segment.

48, equivalent to 12 long and 9 high, can be added.

I don't even know, your teacher is too good!

The first radius is three, the diameter is six, Xiao Ming, that one is fifty circles, and the last one is ninety, and I won't do anything else.

It's okay to set formulas, read books or look up formulas if you can't memorize them, and the numbers are set in.

It's all pretty simple. Give you the answer directly, you won't next time, figure it out for yourself.

Arc length formula. Circumferential formula.

1. The unit price ratio is 6 :5 = :1 2 x=

Apple: Yuan Pear: Yuan.

2: The ratio of technology to composition is 9:10, and the ratio of composition to mathematics is 5:7, and they are multiplied by 2 to get 10:14, so that the three ratios are 9:10:14That's easy to calculate.

99 (9+10+14) 14-99 (9+10+14) 10=42-30=12 (person).

The first question, the weight of the fruit * unit price is the money spent, because it is a ratio, so let x apple is 6 * 2x = 12x pear is 5 * 3x = 15x 12x + 15x = 27x because this total ** is 18 yuan.

So 27x=18x=

So apples are equal to 12* pears and 15*

Question 2 The ratio of the number of people in the science group to the composition group is 9:10 Set the number of people in the science group 9x Composition group 10x The ratio of the number of people in the composition group to the mathematics group is 5:7 Because the composition group has already been set, the mathematics group is 14x (both are doubled, multiplied by two).

9x+10x+14x=99 x=3 Composition is 30 and Math is 42

If the unit price ratio of apples and pears is 6:5 and the weight ratio of apples and pears is 2:3, the total price ratio of apples and pears bought by Honghong is (6 2):

5 3) = 12:15, that is, 18 yuan is divided into 27 equal parts, the total price of apples accounts for 12 parts, and pears account for 15 parts, so apples cost 8 yuan, and pears cost 10 yuan.

Have you learned to set up unknowns? It's easy to learn!

If there are x people in the composition group, then the science group (9 10) x people, and the mathematics group (7 5) x people in the column equation x+(9 10)x+(7 5)x=99 are solved, and x=30 is obtained

Therefore, there are 42 students in the mathematics group, which is 12 more than in the composition group.

1. Let's say Honghong bought 2Y units of apples and 3Y units of pears, and then assume that each unit of apples is 6a yuan and each unit of pears is 5a yuan.

2y＊6a+3y＊5a=18

The solution is y=(2 3a).

Apple bought: 2y*6a=2*(2 3a) 6a=8 yuan, pears bought: 3y 5a=3*(2 3a)*5a=10 yuan 2, (to be continued).

Proportional conversion questions need to find a benchmark quantity.

In the first question, the unit price of pears is 1, and the unit price of apples is multiplied by the amount respectively, and the ratio of the money obtained is 3, and the ratio of 18 divided by one copy is reversed.

The second question is based on 1 in the composition group, 99 in the science and technology group, and 99 in the math group, and the number of people in the composition group is 30, and then you can do it.

The unit price ratio of apples and pears is 6:5, and the weight ratio of apples and pears bought by Honghong is 2:3, which costs a total of 18 yuan.

Q: How much did saffron spend on apples and how much did it cost to buy pears?

Solution: The amount of money spent to buy apples: The amount of money spent to buy pear blossoms = (6:5) (2:3) = 12:15 = 4:5

The amount of money spent to buy apples: 18 4 (4+5)=18 4 9=8 (yuan).

The amount of money to buy pear flowers: 18-8=10 (yuan).

A: Honghong spent 8 yuan to buy apples and 10 yuan to buy pears.

The ratio of the number of people in the science and technology group to the composition group is 9:10, the ratio of the number of people in the composition group to the mathematics group is 5:7, and it is known that there are 99 people in the three groups.

Solution: The ratio of the number of students in the science group to the composition group is 9:10

The ratio of students in the composition group to the math group is 5:7=10:14

The ratio of students in the Technology, Composition, and Math groups is 9:10:14

There are 99 (14-10) (9+10+14)=99 4 33=12 (people) in the math group than in the essay group

A: There are 12 more students in the Mathematics group than in the Composition group.

1.Original formula: (m+1) x 2-2mx+m-2=0(1):

2m) 2-4(m+1)(m-2) 0(2):(2m) 2-4(m+1)(m-2) 0 is b 2-4ac2: if the bottom edge is 2a, then the waist is 4a, and the bottom edge is high.

15a，1/2*2a*

15a=415 find a

(1) Square the known conditions.

2) 2x the angle of cos?

3) Synthesis of the first two.

sina + cosa = 3 2, we can see that a is the first quadrant angle.

But sina + cosa = 2sin(a+ 4) 2<3 2

There is no real solution to this problem.